What is the vertex form of #y= 4x^2-5x-1 #?

Question

What is the vertex form of #y= 4x^2-5x-1   #?

Hazel Anglin · Accepted Answer

The vertex form is: #y=4(x-5/8)^2-41/16#.

Refer to the explanation for the process.

#y=4x^2-5x-1# is a quadratic formula in standard form: #ax^2+bx+c#, where: #a=4#, #b=-5#, and #c=-1# The vertex form of a quadratic equation is: #y=a(x-h)^2+k#, where: #h# is the axis of symmetry and #(h,k)# is the vertex. The line #x=h# is the axis of symmetry. Calculate #(h)# according to the following formula, using values from the standard form: #h=(-b)/(2a)# #h=(-(-5))/(2*4)# #h=5/8# Substitute #k# for #y#, and insert the value of #h# for #x# in the standard form. #k=4(5/8)^2-5(5/8)-1# Simplify. #k=4(25/64)-25/8-1# Simplify. #k=100/64-25/8-1# Multiply #-25/8# and #-1# by an equivalent fraction that will make their denominators #64#. #k=100/64-25/8(8/8)-1xx64/64# #k=100/64-200/64-64/64# Combine the numerators over the denominator. #k=(100-200-64)/64# #k=-164/64# Reduce the fraction by dividing the numerator and denominator by #4#. #k=(-164-:4)/(64-:)# #k=-41/16# Summary #h=5/8# #k=-41/16# Vertex Form #y=4(x-5/8)^2-41/16# graph{y=4x^2-5x-1 [-10, 10, -5, 5]}

Kennedy Adams · Answer

undefined The vertex form of the quadratic equation $ y = 4x^2 - 5x - 1 $ is $ y = 4(x - \frac{5}{8})^2 - \frac{41}{8} $.